[HN Gopher] The math behind mind-reading tricks
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The math behind mind-reading tricks
Author : theafh
Score : 84 points
Date : 2022-05-27 14:12 UTC (8 hours ago)
(HTM) web link (www.quantamagazine.org)
(TXT) w3m dump (www.quantamagazine.org)
| bern4444 wrote:
| My own fun math mind reading trick:
| https://sambernheim.com/blog/the-party-math-trick
| kick_in_the_dor wrote:
| I came here for magic tricks and was disappointed.
| curiousgal wrote:
| What is your partner's age? Add two zeros to the end. Minus
| your birth year from that number. Add the current year to this
| equation. The result should be yours and your partner's current
| ages.
| Someone wrote:
| Not if the one asked to do that calculation hasn't had their
| birthday yet this year (in which case "current year - birth
| year" doesn't yield age in years) or their birth year was
| over 99 years ago (in which case adding "current year - birth
| year" overflows into the partner's age)
| neogodless wrote:
| Close but not quite. If it helps, both birthdays have not
| occurred yet this year. If I assume BOTH have, the trick
| works.
| aaaaaaaaaaab wrote:
| Here's a trick I came up with: 1. think of any number! 2. Add
| your age to the number! 3. Subtract the original number!
|
| _magical hand motions_
|
| The result is your age!
| nestorD wrote:
| Fun fact, about a hundred years ago, some pseudo-mediums would
| use a similar kind of math trick to "guess" the number of
| _deceased_ brother and sisters of someone coming to see them in
| order to do a Seance.
|
| Morality aside, discovering that highly surprised me because I
| always thought that those techniques were transparent to all
| people. But I guess math literacy in the US wasn't always
| great...
| Swizec wrote:
| Statistics and probabilities are incredibly unintuitive to
| people. That's why it's called The Birthday Paradox and not the
| birthday obvious result
|
| > The birthday paradox is that, counterintuitively, the
| probability of a shared birthday exceeds 50% in a group of only
| 23 people.
|
| https://en.m.wikipedia.org/wiki/Birthday_problem
| WhitneyLand wrote:
| They are deceptively unintuitive also. I went back to the
| Monty Hall problem more than once after I previously thought
| I had it figured out:
|
| Suppose you're on a game show, and given the choice of three
| doors: Behind one door is a car; behind the others, goats.
| You pick a door, say No. 1, and the host, who knows what's
| behind the doors, opens another door, say No. 3, which has a
| goat. He then says to you, "Do you want to pick door No. 2?"
| Is it to your advantage to switch your choice?
|
| https://en.wikipedia.org/wiki/Monty_Hall_problem
| pchristensen wrote:
| Here's how I was able to understand it. There are 100
| doors, you choose 1, then host opens 98 doors with goats.
| Do you keep your door or take the 1 remaining?
|
| In the original, the 3 doors muddles the odds. But with
| 100, you're choosing between the 1 you chose on the first
| try, or the 99 you didn't choose.
| jwilk wrote:
| This doesn't help my intuition at all.
| derriz wrote:
| You're actually being given the choice between "my
| initial pick contains the prize" or "the prize is
| contained in any of the other 99 boxes". Okay - the host
| does a routine to add to the "drama" by revealing 98 non-
| prize boxes but since he will never reveal the prize
| during this process, he isn't giving you any more
| information.
| quesera wrote:
| This is how I think about it:
|
| * Your first choice had a natural chance of 1% to be
| correct.
|
| * After eliminating the 98 other options, a random choice
| of the remaining two is 50% likely to be correct.
|
| * BUT, you know that the _reason_ the door you picked was
| not opened is because you picked it! Its inclusion in the
| set-of-two is _not_ random.
|
| * So your first choice is isn't any more likely to be
| correct that it was when you first you chose it.
| patrickdavey wrote:
| The way I think about it (which I hope is correct): when
| you first chose your probability of guessing correctly
| was 1 in 100. It doesn't magically become 1 in 2 when the
| other 98 doors are opened.
| lijogdfljk wrote:
| Funny, i think (but am not arguing i'm correct) it _does_
| change probability when the doors change. As i
| explained[1] in the sister post.
|
| Perhaps this is some mathematical concept? To me i'm
| viewing every choice as a dice roll. I have no question
| that the first dice roll was worse than the 2nd. However
| to me the 2nd is a die with two faces. Two choices, of
| which both are equally possible.
|
| To look at it differently. The doors changing probability
| to me sounds like.. imagine two people, the PersonA has
| this primary scenario. They had 100 doors, chose one, and
| are then asked if they'd like to switch. PersonB then is
| asked the second question, of the 2 doors, which do they
| want? PersonA and B are standing next to each other. The
| doors are the same for both of them. Why would PersonA
| have different odds than PersonB? And how could PersonB's
| odds be any different than 50/50 with no prior history of
| the doors?
|
| Is there some fundamental difference between the grand
| idea of probability and "reality" as i'm trying to
| describe it here?
|
| [1]: https://news.ycombinator.com/item?id=31535557
| lijogdfljk wrote:
| Man, i still don't get it. Well, maybe i do? :shrug:
|
| The way i see it, the other door has a higher chance of
| being correct than the door you _originally_ chose.
| However the two are fundamentally different, are they
| not?
|
| In reality both doors are equally likely to be correct,
| no? So yes, changing would be better compared to your
| original answer - but "to change or not" evaluates both
| doors in my mind, and each is a 50% chance, no?
|
| Which i _think_ is what you 're saying here:
|
| > So your first choice is isn't any more likely to be
| correct that it was when you first you chose it.
|
| But nevertheless the problem has me a bit confused in
| what is expected to be "right". It seems to me there is
| no right, it is a 50/50, unless you are strictly
| answering the moot point of "Which is better? The door
| you first chose, or the potential other door?".
|
| Ie i guess i view the original choice as irrelevant the
| moment the variables change. I feel i misunderstand the
| goal of the "problem".
| vitus wrote:
| Or, an even more telling variation:
|
| You pick door #1.
|
| The host goes and, in order, one by one, opens doors
| 2-52, conspicuously skips door 53, and then opens 54-100.
| Would you like to stick with your original choice?
| cgriswald wrote:
| I think this variation makes it more intuitively clear
| one should switch in the case with 100 doors, but I think
| it makes the explanation of _why_ less clear because of
| the extra, unnecessary information.
| criddell wrote:
| I've never been able to wrap my head around this one.
|
| If in a group 23 people at least two share a birthday with a
| ~50% chance, would that mean if I have a group of 22 people
| and pick a random day of the year, there's also a ~50% chance
| of somebody having a birthday on that day?
| erehweb wrote:
| No. There's about a 22/365 chance that somebody has a
| birthday on that day. But this is getting close to the
| intuition. Go through the 23 people. For each one of them,
| there's about a 22/365 somebody else in the group has the
| same birthday as they do. So adding these probabilities up
| over all 23, you get 23 * 22/365. Of course, adding up the
| probabilities is obviously wrong since the events are not
| independent (and 23 * 22/365 > 1) but it gives the
| intuition that the probability of a shared birthday grows
| like N^2. [and actually, you can add up expectations]
| [Edits for correctness]
| [deleted]
| [deleted]
| whatever1 wrote:
| For all these problems I find it easier to just simulate it
| and plot the results. Do a loop with 100k iterations. In
| each iteration generate 23 numbers between 1,365 from a
| uniform distribution. Then check in how many iterations you
| had duplicate entries.
|
| You will see that asymptomatically you will reach 50k cases
| conanite wrote:
| Covid cases are asymptomatically approaching zero.
| However, your duplicate entries will reach their target
| _asymptotically_ :)
|
| https://en.wikipedia.org/wiki/Asymptote
| whatever1 wrote:
| I know, Apple keyboard does not.
| dhosek wrote:
| I think the easiest way to think about it is to invert the
| problem. What is the probability that no one shares a
| birthday?
|
| One person, nobody shares a birthday because they're alone.
|
| Two people, it's going to be a 364/365 [?] 99.7% that they
| don't share a birthday.1
|
| Now add a third person. We have the initial probability for
| the first pair and then the third person can't share a
| birthday with either of them, so the probability is going
| to be 363/365 that they don't share a birthday with either,
| but since we have two events that both must occur, we
| multiply the probabilities. That gives us (363 _364) /3652
| [?] 99.2%
|
| For the fourth person it becomes (362_363 _364) /3653 [?]
| 98.4%
|
| You'll notice that the probability is decreasing more and
| more with each step.2
|
| In general, for _n* people we would have3 _P_ = 365!
| /((365- _n_!)365n). At _n_ = 23, we drop below 50%.
|
| This technique is useful for a lot of statistical
| calculations, e.g., for calculating the probabilities of
| poker hands, some of them are easier to calculate the
| probability of not getting the hand then the probability of
| getting the hand (a pair being the most obvious case).
|
| [?]
|
| 1. We're mostly programmers here so we're going to pretend
| leap years don't exist.
|
| 2. This will not continue indefinitely, obviously. The
| first derivative hits a minimum4 around _n_ = 20 for this
| formula and at _n=365_. The graph kind of looks like cos th
| over the range [0,p] but not exactly, it's just the first
| well-known graph that comes to mind of this shape.
|
| 3. If you're eagle eyed, you might notice that my examples
| before had a denominator of 365 raised to the _n_ -1 power.
| Then you'll also notice that my factorials have _n_ terms
| and end with 365 and not 364. Remember our lonely birthday
| dude? His probability of not sharing a birthday (1) can be
| thought of as 365 /365 and should also be part of the
| product, but when we're doing the math for small numbers,
| it's easier to just ignore that but for a general formula,
| keeping him in the product makes the formula clearer.
|
| 4. Minimum because we're _decreasing_ , remember.
| bena wrote:
| Statistics aren't reflective. They're aggregate.
|
| It's not a 50% chance of being on the date you chose, but
| that any two dates are the same.
|
| If you pick a random date, Alice has a 364/365 chance of
| _not_ having a birthday that day. Bob has a 363 /365 chance
| of not having a birthday on that day or on Alice's birthday
| independently of whether or not Alice's birthday is on the
| random date. So we take Alice's chances and Bob's chances
| and we can multiply them together to get our odds that none
| of these people share a date. And of course, if we know the
| odds of something not happening, the odds of it happening
| is the opposite.
| munificent wrote:
| The trick is to realize that "same birthday" isn't a
| property of _people_ , it's a question about _pairs_ of
| people.
|
| A group of 23 people has 253 unique pairs. So if I ask, "In
| a room with 253 pairs of people, what are the odds that in
| one of those pairs, both have the same birthday?" Now
| around 50% seems a lot more intuitive.
|
| It's unintuitive because the number of pairs increases
| quadratically with the number of people.
| dwighttk wrote:
| that's just 1 date that has to be matched, so you get a
| lower probability of matching, comparing 22 dates to 1...
| in the 23 person scenario, you have 23 dates that can match
| with any of the 23 dates.
| andi999 wrote:
| No. The chance would be much smaller. If you choose 22
| random days though...
| criddell wrote:
| Your comment and erehweb's finally made it click for me.
| Thanks.
| charcircuit wrote:
| I was expecting an article talking about probability instead of
| algebra. People are bad at coming up with random numbers
| especially if you put some restrictions on what number they are
| allowed to generate.
| vincnetas wrote:
| show me a kid that can multiply by 13 tree digit number in theyr
| head and that would be enough magic trick for me :)
| cgriswald wrote:
| Define 'kid.' By middle school, at least, I could have done
| that even without the (10x + 3x) simplification.
|
| The real trick would be finding a three digit number that can
| be multiplied by 77 and still be a three digit number.
| slingnow wrote:
| I don't think any one of these could be classified as a "mind-
| reading magic trick".
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